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4 years ago

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When the obligations of a partnership can’t be met, each partner is liable for the obligation. This characteristic is called A.

limited life. B. unlimited liability. C. limited liability. D. mutual agreement

1 answer:

Anna11 [10]4 years ago

4 0

The answer to this question is <span>B. unlimited liability
In partnership, all partners involved acted as one united entity , even though the percentage of ownership is different.
So, when a partner make a certain debt in the name of the firm, all partners in that firm is required to a certain percentage of that debt depending on how much ownership they had in the firm. If one partner couldn't pay their percentage, other members are legally binded to chip in.</span>

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Solve each system of equations using substitution. Show all work.

Ahat [919]

It would be so much easier with elimination but ok
1.
x+y=1
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2.
x-2y=2
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x=2y+2
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10y+10-3y=-11
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3.
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(5/2,-1/2)

1. (-3,4)
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Convert the Cartesian equation x^2 + y^2 = 16 to a polar equation.

RideAnS [48]

Answer:

Problem 1: $r=4$

Problem 2: $r=-2\sin(\theta)$

Problem 3: $r\sin(\theta)=3$

Step-by-step explanation:

Problem 1:

So we are going to use the following to help us:

$x=r \cos(\theta)$

$y=r \sin(\theta)$

$\frac{y}{x}=\tan(\theta)$

So if we make those substitution into the first equation we get:

$x^2+y^2=16$

$(r\cos(\theta))^2+r\sin(\theta))^2=16$

$r^2\cos^2(\theta)+r^2\sin^2(\theta)=16$

Factor the $r^2$ out:

$r^2(\cos^2(\theta)+\sin^2(\theta))=16$

The following is a Pythagorean Identity: $\cos^2(\theta)+\sin^2(\theta)=1$ .

We will apply this identity now:

$r^2=16$

This implies:

$r=4 \text{ or } r=-4$

We don't need both because both of include points with radius 4.

Problem 2:

$x^2+y^2+2y=0$

$(r\cos(\theta))^2+(r\sin(\theta))^2+2(r\sin(\theta))=0$

$r^2\cos^2(\theta)+r^2\sin^2(\theta)+2r\sin(theta)=0$

Factoring out $r^2$ from first two terms:

$r^2(\cos^2(\theta)+\sin^2(\theta))+2r\sin(\theta)=0$

Apply the Pythagorean Identity I mentioned above from problem 1:

$r^2(1)+2r\sin(\theta)=0$

$r^2+2r\sin(\theta)=0$

or if we factor out r:

$r(r+2\sin(\theta))=0$

$r=0 \text{ or } r=-2\sin(\theta)$

r=0 is actually included in the other equation since when theta=0, r=0.

Problem 3:

$y=3$

$r\sin(\theta)=3$

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